| New Reply |
Computing Matrix, finding kernel and image |
Share Thread | Thread Tools |
| Nov2-10, 11:22 AM | #1 |
|
|
Computing Matrix, finding kernel and image
Let T: R[x]2[tex]\rightarrow[/tex] R[x]3 be defined by T(P(x))=xP(x). Compute the matrix of x with respect to bases {1,x,x2} and {1,x,x2,x3}. Find the kernel and image of T.
I know how to do this when given bases without exponents, however I do not know exactly what this is saying and therefore am having a hard time starting it. |
| Nov2-10, 11:59 AM | #2 |
|
|
Let's compute the first column.
You take the first basis element, that is 1. Now you'll have to express T(1)=x in terms of the basis {1,x,x²,x³}. That would be (0,1,0,0). So the first column would consist out of [tex] \left(\begin{array}{ccc} 0 & ? & ? \\ 1 & ? & ? \\ 0 & ? & ? \\ 0 & ? & ? \end{array} \right) [/tex] Now for the second and third column, you'll have to express T(x) and T(x²) in terms of the basis {1,x,x²,x³}. |
| Nov2-10, 03:15 PM | #3 |
|
|
Okay so matrix T(x)=x2 and with respect to the basis {1,x,x²,x³} the second column would be (0,0,1,0), T(x2)=x3 and with respect to the basis {1,x,x²,x³} would be (0,0,0,1).
I would then say that T=(0 0 0) (1 0 0) (0 1 0) (0 0 1) with respect to {1,x,x²,x³} right? Not with respect to {1,x,x²} and {1,x,x²,x³} so then to find the im(T) I just use (0 0 0) (1 0 0) (0 1 0) (0 0 1) and solve it with respect to the basis {1,x,x²,x³} right? Then how would I go about solving for the ker(T)? |
| Nov2-10, 03:24 PM | #4 |
|
|
Computing Matrix, finding kernel and image
For the Ker(T). Take your matrix A. Then wonder when Ax=0. This you should be able to solve easily (it's a system with 3 unknowns and 4 equations)
|
| Nov2-10, 03:34 PM | #5 |
|
|
So, if (x,y,z) is in the kernel, then you must have
[tex] \left( \begin{array}{ccc} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x\\ y\\ z \end{array} \right) = 0 [/tex] It's not hard to see that this can only be the case iff x=y=z=0 |
| Nov2-10, 03:45 PM | #6 |
|
|
right, that just seemed way too easy I thought I was doing it wrong.
Then the im(T) is the span of e2, e3, and e4 |
| Nov2-10, 03:47 PM | #7 |
|
|
Yes, your image is correct to.
And remember, math doesnt have to be difficult
|
| New Reply |
| Tags |
| bases, image, kernel, linear algebra |
| Thread Tools | |
Similar Threads for: Computing Matrix, finding kernel and image
|
||||
| Thread | Forum | Replies | ||
| Matrix Image and Kernel | Calculus & Beyond Homework | 4 | ||
| Basis for Image and Kernel of matrix | Calculus & Beyond Homework | 3 | ||
| Finding a Matrix whose kernel is spanned by 2 vectors | Calculus & Beyond Homework | 3 | ||
| Finding kernel of matrix transformation | Calculus & Beyond Homework | 2 | ||
| [SOLVED] Kernel and image of a matrix A | Calculus & Beyond Homework | 4 | ||